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91Ó°ÊÓ

When are the samples considered large enough for the sampling distribution of the difference between two sample proportions to be (approximately) normal?

Short Answer

Expert verified
The samples are considered large enough for the sampling distribution of the difference between two sample proportions to be approximately normal when the expected number of successes (np) and failures (n(1-p)) in both groups are greater than or equal to 5.

Step by step solution

01

Understand the concept of Sample Distribution

The sampling distribution of a statistic is the distribution of all possible values of the statistic for all possible samples from the same population of a given size. The Central Limit Theorem (CLT) states that the distribution of sample means approximates a normal distribution (that is, gets closer to a normal distribution) as the sample size increases.
02

Fundamental condition for Sample Proportions

For sample proportions to be considered large enough, the sample size (n) should be sufficiently large as to satisfy two basic conditions – the expected number of successes, np, and failures, n(1-p), are both greater than or equal to 5.
03

Applying the criteria

In order to consider the samples large enough for the difference between the proportions to be approximately normal, you should ensure that the conditions, np>5 and n(1-p)>5, for both groups. If these conditions are met, you can say that the sample is large enough.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Central Limit Theorem
The Central Limit Theorem (CLT) is a key concept in statistics. It explains why we often see the normal distribution in statistics, even when the data itself doesn't seem normal at first glance. Here's what the theorem says: as the sample size grows, the distribution of the sample mean will become approximately normal. This means that even if you're sampling from a population that is not normally distributed, the average of your sample will tend to form a normal distribution (bell-curve shape) if you have a sufficiently large sample size.

The CLT is powerful because it allows statisticians to make inferences about population parameters. Since the sample mean will approximate a normal distribution, we can use normal distribution characteristics to predict probabilities and calculate margins of error. This is why you will often hear that a sample size of 30 is large enough for the CLT to apply, although, in some cases, fewer observations may work as well. The actual needed size depends on how the data was distributed and the shape of the original data. And once sample sizes are large enough, our predictions and calculations become more reliable.
Sample Proportions
Sample proportions are a tool we use to understand part of a whole population. They can be thought of as the fraction or percentage of a sample that displays a particular attribute. For example, if you survey 100 people and 55 said they prefer tea over coffee, your sample proportion for tea preference is 0.55 or 55%. Sample proportions come into play especially when making predictions or drawing conclusions about a whole population based on samples. To trust these predictions, it is crucial that the sample size is adequately large. The reliability of sample proportions is anchored on specific conditions.
For a sample proportion to approximate a normal distribution, certain conditions must be met. These conditions require that the expected number of successes (np) and failures (n(1-p)) are both greater than or equal to 5. This ensures that the sample size is "large enough" to rely on the approximation offered by the normal distribution. If these criteria are not satisfied, any conclusions drawn from the sample may be inaccurate.
Normal Distribution
The normal distribution, often referred to as the bell curve, is a common probability distribution in statistics. It is characterized by its symmetric shape, where most of the data points cluster around the mean, and the probability of values decreases as you move further away from the mean. Understanding the normal distribution is crucial because many statistical tests rely on it. It serves as a cornerstone for making inferences about data. For large sample sizes, the normal distribution allows statisticians to conduct hypothesis testing and construct confidence intervals, among other things. One of the reasons the normal distribution is so pervasive is because of the Central Limit Theorem. As long as your samples are large enough, the distribution of the sample means will tend to be normal, regardless of the distribution of the original population. This property makes the normal distribution a useful tool for interpreting data that might otherwise be too complex or varied to analyze directly. Hence, when sample data fits a normal distribution, predictions and conclusions can be more confidently made.
Sample Size
Sample size refers to the number of observations in a sample, which plays a critical role in statistical analysis and interpretations. The size can significantly affect the accuracy of the sample in representing the whole population. To determine the sampling distribution's shape, especially in contexts like estimating population parameters through sampling, the sample size is crucial. For the sample mean or sample proportion to approximate normal distribution, the sample size must be large enough. Typically, when discussing sample proportions, the sample size is "large enough" if both the products of the sample size and the probability of success (np) and failure (n(1-p)) are greater than 5. A larger sample size reduces the margin of error, providing more reliable estimates of population parameters. However, it's important to balance the desire for larger sample sizes with practicality since very large samples can be costly and time-consuming to collect. Thus, statisticians aim to find an optimal sample size that ensures precision without incurring unnecessary data collection costs.

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Most popular questions from this chapter

Employees of a large corporation are concerned about the declining quality of medical services provided by their group health insurance. A random sample of 100 office visits by employees of this corporation to primary care physicians during 2004 found that the doctors spent an average of 19 minutes with each patient. This year a random sample of 108 such visits showed that doctors spent an average of \(15.5\) minutes with each patient. Assume that the standard deviations for the two populations are \(2.7\) and \(2.1\) minutes, respectively. a. Construct a \(95 \%\) confidence interval for the difference between the two population means for these two years. b. Using a \(2.5 \%\) level of significance, can you conclude that the mean time spent by doctors with each patient is lower for this year than for \(2004 ?\) c. What would your decision be in part \(\mathrm{b}\) if the probability of making a Type I error were zero? Explain.

According to a report in The New York Times, in the United States, accountants and auditors earn an average of \(\$ 70,130\) a year and loan officers carn \(\$ 67,960\) a year (Jessica Silver-Greenberg, The New York Times, April 22,2012 ). Suppose that these estimates are based on random samples of 1650 accountants and auditors and 1820 loan officers. Further assume that the sample standard deviations of the salaries of the two groups are \(\$ 14,400\) and \(\$ 13,600\), respectively, and the population standard deviations are equal for the two groups. a. Construct a \(98 \%\) confidence interval for the difference in the mean salaries of the two groupsaccountants and auditors, and loan officers. b. Using a \(1 \%\) significance level, can you conclude that the average salary of accountants and auditors is higher than that of loan officers?

A state that requires periodic emission tests of cars operates two emission test stations, \(\mathrm{A}\) and \(\mathrm{B}\), in one of its towns. Car owners have complained of lack of uniformity of procedures at the two stations, resulting in different failure rates. A sample of 400 cars at Station A showed that 53 of those failed the test; a sample of 470 cars at Station \(\mathrm{B}\) found that 51 of those failed the test. a. What is the point estimate of the difference between the two population proportions? b. Construct a 95\% confidence interval for the difference between the two population proportions. c. Testing at a \(5 \%\) significance level, can you conclude that the two population proportions are different? Use both the critical-value and the \(p\) -value approaches.

Refer to Exercise \(10.32\). As mentioned in that exercise, according to the credit rating agency Equifax, credit limits on newly issued credit cards increased between January 2011 and May 2011 . Suppose that random samples of 400 new credit cards issued in January 2011 and 500 new credit cards issued in May 2011 had average credit limits of \(\$ 2635\) and \(\$ 2887\), respectively. Suppose that the sample standard deviations for these two samples were \(\$ 365\) and \(\$ 412\), respectively. Now assume that the population standard deviations for the two populations are unknown and not equal. a. Let \(\mu_{1}\) and \(\mu_{2}\) be the average credit limits on all credit cards issued in January 2011 and in May 2011 , respectively. What is the point estimate of \(\mu_{1}-\mu_{2}\) ? b. Construct a \(98 \%\) confidence interval for \(\mu_{1}-\mu_{2}\). c. Using a \(1 \%\) significance level, can you conclude that the average credit limit for all new credit cards issued in January 2011 was lower than the corresponding average for all credit cards issued in May 2011 ? Use both the \(p\) -value and the critical-value approaches to make this test.

As mentioned in Exercise \(10.26\), a town that recently started a single-stream recycling program provided 60 -gallon recycling bins to 25 randomly selected households and 75 -gallon recycling bins to 22 randomly selected households. The average total volumes of recycling over a 10 -week period were 382 and 415 gallons for the two groups, respectively, with standard deviations of \(52.5\) and \(43.8\) gallons, respectively. Suppose that the standard deviations for the two populations are not equal. a. Construct a \(98 \%\) confidence interval for the difference in the mean volumes of 10 -week recyclying for the households with the 60 - and 75 -gallon bins. b. Using a \(2 \%\) significance level, can you conclude that the average 10 -week recycling volume of all households having 60 -gallon containers is different from the average 10 -week recycling volume of all households that have 75 -gallon containers? c. Suppose that the sample standard deviations were \(59.3\) and \(33.8\) gallons, respectively. Redo parts a and b. Discuss any changes in the results.

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